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受欢迎的三角函数 >

tan(x)=tan(2x)

解答

tan (x) = tan (2 x)

解答

x = πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n

求解步骤

tan (x) = tan (2 x)

两边减去

tan (2 x)

tan (x) - tan (2 x) = 0

使用三角恒等式改写

- tan (2 x) + tan (x)

使用倍角公式:

tan (2 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

= - \frac{2 tan ( x )}{1 - tan ^{2} ( x )} + tan (x)

化简

- \frac{2 tan ( x )}{1 - tan ^{2} ( x )} + tan (x) : \frac{- tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )}

- \frac{2 tan ( x )}{1 - tan ^{2} ( x )} + tan (x)

将项转换为分式:

tan (x) = \frac{tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

= - \frac{2 tan ( x )}{1 - tan ^{2} ( x )} + \frac{tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 2 tan ( x ) + tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

乘开

- 2 tan (x) + tan (x) (1 - tan^{2} (x)) : - tan (x) - tan^{3} (x)

- 2 tan (x) + tan (x) (1 - tan^{2} (x))

乘开

tan (x) (1 - tan^{2} (x)) : tan (x) - tan^{3} (x)

tan (x) (1 - tan^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = tan (x), b = 1, c = tan^{2} (x)

= tan (x) \cdot 1 - tan (x) tan^{2} (x)

= 1 \cdot tan (x) - tan^{2} (x) tan (x)

化简

1 \cdot tan (x) - tan^{2} (x) tan (x) : tan (x) - tan^{3} (x)

1 \cdot tan (x) - tan^{2} (x) tan (x)

1 \cdot tan (x) = tan (x)

1 \cdot tan (x)

乘以:

1 \cdot tan (x) = tan (x)

= tan (x)

tan^{2} (x) tan (x) = tan^{3} (x)

tan^{2} (x) tan (x)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

tan^{2} (x) tan (x) = tan^{2 + 1} (x)

= tan^{2 + 1} (x)

数字相加:

2 + 1 = 3

= tan^{3} (x)

= tan (x) - tan^{3} (x)

= tan (x) - tan^{3} (x)

= - 2 tan (x) + tan (x) - tan^{3} (x)

同类项相加:

- 2 tan (x) + tan (x) = - tan (x)

= - tan (x) - tan^{3} (x)

= \frac{- tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )}

= \frac{- tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )}

\frac{- tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )} = 0

用替代法求解

\frac{- tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )} = 0

令

: tan (x) = u

\frac{- u - u ^{3}}{1 - u ^{2}} = 0

\frac{- u - u ^{3}}{1 - u ^{2}} = 0 : u = 0, u = i, u = - i

\frac{- u - u ^{3}}{1 - u ^{2}} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- u - u^{3} = 0

解

- u - u^{3} = 0 : u = 0, u = i, u = - i

- u - u^{3} = 0

因式分解

- u - u^{3} : - u (u^{2} + 1)

- u - u^{3}

使用指数法则:

a^{b + c} = a^{b} a^{c}

u^{3} = u^{2} u

= - u^{2} u - u

因式分解出通项

- u

= - u (u^{2} + 1)

- u (u^{2} + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or u^{2} + 1 = 0

解

u^{2} + 1 = 0 : u = i, u = - i

u^{2} + 1 = 0

将

1

到右边

u^{2} + 1 = 0

两边减去

1

u^{2} + 1 - 1 = 0 - 1

化简

u^{2} = - 1

u^{2} = - 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = - 1, u = - - 1

化简

- 1 : i

- 1

使用虚数运算法则:

- 1 = i

= i

化简

- - 1 : - i

- - 1

使用虚数运算法则:

- 1 = i

= - i

u = i, u = - i

解为

u = 0, u = i, u = - i

u = 0, u = i, u = - i

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

\frac{- u - u ^{3}}{1 - u ^{2}}

的分母，令其等于零

解

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

1 - u^{2} - 1 = 0 - 1

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

以下点无定义

u = 1, u = - 1

将不在定义域的点与解相综合:

u = 0, u = i, u = - i

u = tan (x)

代回

tan (x) = 0, tan (x) = i, tan (x) = - i

tan (x) = 0, tan (x) = i, tan (x) = - i

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 0

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = i :

无解

tan (x) = i

无解

tan (x) = - i :

无解

tan (x) = - i

无解

合并所有解

x = πn

作图

流行的例子

2 cos (x) = \frac{1}{2}

cos (x) + 0.65 = 0

sin^{2} (x) = 0.12

cos (x) = \frac{7}{12}

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4 tan (x) + 2 sin (x) cos (x) = 0